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Question
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
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Solution
LHS = `(cot^2θ(secθ - 1))/((1 + sinθ)) `
= `(cot^2θ(secθ - 1)(1 - sinθ)(secθ + 1))/((1 + sinθ)(1 - sinθ)(secθ + 1))`
= `(cot^2θ(secθ - 1)(secθ + 1)(1 - sinθ))/((1 + sinθ)(1 - sinθ)(secθ + 1))`
= `(cot^2θ(sec^2θ - 1)(1 - sinθ))/((1 - sin^2θ)(1 + secθ))`
= `(cot^2θ(tan^2θ)(1 - sinθ))/((cos^2θ)(1 + secθ))` (∵ `tan^2θ = sec^2θ - 1,1 - sin^2θ = cos^2θ`)
= `((cotθtanθ)^2(1 - sinθ))/((cos^2θ)(1 + secθ))`
= `(1(1 - sinθ))/((cos^2θ)(1 + secθ))` (∵ cotθtanθ = 1)
= `sec^2θ((1 - sinθ)/(1 + secθ))`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
